How would an object need to move in order for total distance traveled and displacement to be equal?
1 answer:
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A) See ray diagram in attachment (-6.0 cm)
By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:
where
q is the distance of the image from the lens
f = -10 cm is the focal length (negative for a diverging lens)
p = 15 cm is the distance of the object from the lens
Solving for q,
B) The image is upright
As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:
where are the size of the image and of the object, respectively.
Since q < 0 and p > o, we have that , which means that the image is upright.
C) The image is virtual
As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.
This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual
To solve this problem it is necessary to apply the concepts related to the Conservation of Energy, for which it is necessary that any decrease made through the potential energy, is equivalent to the gain given in the kinetic energy or vice versa.
Mathematically this can be expressed as
Since there is no final potential energy (the height is zero), and the initial potential energy is equivalent to the work done we have to
Therefore the non-conservative work was done on the boy is 1.4kJ